continuity and $C^2$ solution of a series

For $\alpha$ with $|\alpha|=2$ let $P$ be a homogenous harmonic Polynom of degree $2$ with $D^\alpha P\ne0$ (e.g. take $P=2x_1x_2$). Choose $\eta\in C^\infty_0(\{x:|x|<2\})$ with $\eta=1$ when $|x|<1$ and $\eta=0$ when $|x|\geq2$, set $t_k=2^k$ and $c_k=\frac1k$ with $\sum c_k$ divergent. Define $f(x)=\sum\limits_0^\infty c_k\Delta(\eta P)(t_kx)$.

How do you proof that $f$ is continuous but that $\Delta u=f$ does not have a $C^2$ solution in any neighborhood of the origin?

Since $P$ is a harmonic polynomial, you have that $\triangle P = 0$. Therefore $$ \triangle(\eta P) = \triangle \eta \cdot P + 2\nabla\eta \nabla P $$ Observe that in particular $$ \operatorname{supp}(\triangle(\eta P)) \subset \{ 1\leq |x| \leq 2\}~. $$ Therefore for every $x \neq 0$, there exists a small neighborhood $V_x$ of $x$ (ball of radius $\frac12 |x|$, say), such that $f(x)$ can be expressed as a finite sum of continuous functions on $V_x$ (that is, for all but finitely many of that $\triangle(\eta P)(t_k x)$ are zero on $V_x$). So $f(x)|_{V_x}$ is a continuous function away from 0. For $x = 0$, observe that the definition requires $f(0) = 0$. It is easy to check that, since $c_k \searrow 0$, and since we have that $f(x) = c_k \triangle(\eta P)(t_k x)$ for $2^{-k} \leq |x| < 2^{-k+1}$, that $f(x) = O\left( \frac{1}{\lvert\log_2 |x|\rvert}\right)$ and hence is continuous at 0.

Now suppose $u$ solve $\triangle u = f$. Away from the origin, define the function $v$ to be $\sum \frac{c_k}{t_k^2} (\eta P)(t_k x)$. Since $|x| > 0$, only finitely many terms are non-zero in the sum (up to $k \approx - \log_2 |x|$). A similar argument as above shows that for any $|x| > 0$ there is a neighborhood of $|x|$ such that $v$ must be twice continuously differentiable, being the sum of a finite number of $C^2$ functions there.

On the other hand, since $t_k$ decreases geometrically, we have that the infinite sum expression defining $v$ is absolutely convergent. And thus $v$ is continuous.

Furthermore, as away from the origin $v$ is given by a finite sum, we can differentiate term by term in the sum. This shows that away from the origin $\triangle v = f$. Hence any continuous solution of $\triangle u = f$ must be $v$ plus some harmonic (and hence $C^\infty$) function.

It suffices to show that the Hessian of $v$ is not continuous. It is here we use the condition that $D^2 P$ does not vanish identically. Let us do it for the case of $P = x_1 x_2$. A direct computation of $\partial^2_{x_1x_2} v$ shows that away from the origin, it is given by

$$ \sum c_k \eta(t_k x) + c_k \partial_1\eta(t_k x) t_kx_1 + c_k \partial_2 \eta(t_k x) t_k x_2 + c_k \partial^2_{12}\eta(t_k x) t_k^2 x_1x_2 $$

The same argument as before shows that the last three terms are only non-zero when $t_k |x| \approx 1$. This means that their contribution to the sum is only $O(c_k) = O(\frac{1}{- \log_2|x|})$, which decays as $x\to 0$. The main first term, however, is bounded below by

$$ \sum_{k < - \log_2 |x|} c_k > \frac12 \ln\left( -\log_2 |x|\right)$$

which diverges as $x\to 0$. In particular, this means that $\partial^2_{1,2}v$ is unbounded as $x\to 0$, showing that the Hessian of $v$ cannot be continuous at the origin.